Triple Integrals in Cylindrical Coordinates — Question 3

Question 3 – Solution

Strategy. Convert both surfaces to functions of $r$, solve for their positive intersection radius, and integrate top minus bottom.

Step 1: Intersection The surfaces are $z=r$ and $z=2-r^2$. Their intersection satisfies $$r=2-r^2\quad\Longleftrightarrow\quad(r-1)(r+2)=0.$$ Since $r\ge 0$, the boundary circle is $r=1$, $z=1$.

Step 2: Integral $$\begin{align*} V&=\int_0^{2\pi}\int_0^1\int_r^{2-r^2}r\,dz\,dr\,d\theta\\ &=2\pi\int_0^1(2-r^2-r)r\,dr =2\pi\left(1-\frac 14-\frac 13\right) =\boxed{\frac{5\pi}{6}}. \end{align*}$$

Verification For $0\le r\le 1$, $2-r^2-r=(1-r)(r+2)\ge 0$, so the upper and lower $z$-bounds never reverse.